The aim of my experiment is to investigate how the mass of potato (concentration of catalase) affects the rate of reaction with hydrogen peroxide.

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AN INVESTIGATION INTO HOW THE MASS OF POTATO AFFECTS THE RATE OF REACTION OF THE ENZYME CATALASE

SKILL AREA P: PLANNING

Aim:

The aim of my experiment is to investigate how the mass of potato (concentration of catalase) affects the rate of reaction with hydrogen peroxide.

Introduction:

In a chemistry lesson I learnt that an enzyme is a biological protein that acts as a catalyst in living cells.  A catalyst is a substance that changes the rate of reaction without being used up.  Therefore it can be re-used.  Catalysts help substances to react more easily by breaking and forming bonds between atoms.  Most catalysts speed up their reactions.  This is because particles need less energy to react, so the reactions proceed more quickly.

The reaction I am investigating occurs in humans because hydrogen peroxide is a by-product of some chemical reactions that take place in our cells. The reason that hydrogen peroxide needs to be broken down is because it is very poisonous as H2O2 but is harmless as water and oxygen.  Here is an equation to show this reaction:

Hydrogen Peroxide                     Oxygen  +  Water

2H2O2                     O2   +  2H2O

However in my investigation I will be using potatoes as it contains living cells and the enzyme catalase.

Key Variables:
In the experiment I could have varied the following factors to find out how the rate of reaction is affected.

Concentration of hydrogen peroxide:
When the concentration of hydrogen peroxide is increased, there are more particles per unit volume.  Therefore, the frequency of collisions increases, and the rate of the reaction increases.  To keep the concentration constant I will use the same concentration throughout the experiment.

Volume of hydrogen peroxide:
If the volume of hydrogen peroxide is large then that means that there is more reacting particles than in a smaller volume of hydrogen peroxide.  Therefore more collisions per second will occur making the reaction fast.  To keep the volume of hydrogen peroxide constant I will use 20ml of hydrogen peroxide in every experiment.

The temperature:
As the temperature increases, the particles gain kinetic energy, and begin to move faster.  As a result, the reacting particles collide more frequently and with greater energy.  Therefore, there are more successful collisions per second, and the rate of the reaction increases.  To keep the temperature of the water bath constant I will use a thermometer to measure the temperature.  I will then add cold water to the water bath if the temperature increases or hot water if the water bath gets cold.

Surface area of the potato:
The surface area of the potato is the area of it, which is exposed to the hydrogen peroxide.  If the potato is cut up into smaller pieces, the surface area of the potato has increases, which means that it can react faster.  This is the same vice versa if the potato is in big pieces, there is a smaller surface area exposed and fewer cut cells therefore few collisions per second of catalase will occur with the hydrogen peroxide.  I will use minced potato to obtain a larger surface area of potato for the reaction to take place.  To keep this constant I will use minced potato throughout my experiment and also chop the potato in the mincing machine for 30seconds in each experiment.

The pH:
Different enzymes work at a certain pH.  They each have their own optimum pH that they work best at.  If the pH is very different to their optimum pH, then the enzymes can be denatured or the rate of reaction can be extremely slow.  This is because the enzyme is not working at its best rate.  To keep the pH of the enzyme constant, I will use the same type of potato and the same enzyme throughout the experiment.

Mass of the potato:
If the mass of the potato is increased this means that the concentration of the catalase is increased.  There are more reacting particles creating more collisions per second, making the rate of the reaction increase.

        In my experiment I am going to vary the mass of the potato. If I decide to vary the mass of the potato I must keep all other factors constant for a fair test. If I didn’t control the concentration of the hydrogen peroxide and used a much more concentrated chemical for the same experiment then there would be many more collisions and a lot more reaction occurring. The same applies to the volume if there is a greater volume of hydrogen peroxide then collisions will occur more frequently. The surface area would affect the experiment greatly as the more surface area there is the more reaction space. All these factors must be controlled to keep the reaction constant. I will use minced potato in my experiment because there will be a higher surface area, therefore more reacting particles will collide.  As catalase are in living cells, if I have minced my potato, the cells will also be cut up releasing the catalase ready to react with the hydrogen peroxide.  This will make the reaction instantaneous.
        The independent variable in my experiment will be the mass of the potato and the dependent variable will be the volume of oxygen produced.

Prediction:
        
I predict that the volume of oxygen produced from the reaction will increase as the mass of potato (catalase) is increased, as there is a greater surface area. Mincing the potato creates this surface area.  If I mince the potato, this means that I will be cutting the potato cells that contain the catalase.  Therefore there will be more catalase exposed to the hydrogen peroxide, making the reaction faster.  As I increase the mass of potato this will increase the concentration of catalase as the enzyme catalase is present in every potato cell.  I think this will happen because if there are more enzyme molecules there will be more of a chance that a hydrogen peroxide molecule will collide with an enzyme molecule. Another reason why I think more gas will be produced with more catalase is because catalase is a catalyst and we have learned in chemistry that catalysts lower the activation energy in an experiment.  I found out in a chemistry lesson that the activation energy is the minimum energy reacting particles must have when they collide if they are to react successfully. 
        
I predict that when the mass of potato is doubled, the rate of the reaction will double.  This is because when the mass of potato is doubled; there will be twice the number of catalase reacting with the same volume of hydrogen peroxide.  Therefore, there will be twice the number of collisions per unit time. Therefore, I predict the graph to be linear and the rate of reaction to be directly proportional to the mass of potato.  However after thinking about my prediction for longer I have realised that when the concentration of catalase equals the concentration of hydrogen peroxide the rate of reaction on the graph will be constant.  Here are diagrams to explain my prediction:

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Preliminary experiment:
        
I carried out preliminary work to see how I would carryout my experiment in my actual plan.  It also helped me on deciding suitable apparatus to use for my real experiment.  A few potatoes were minced using a mincing machine.  5g of minced potato were then obtained using a balance.  I transferred this into a conical flask but with real difficulty.  The problem was that I could not get all of the minced potato into the conical flask without leaving at least 0.1g of potato on the balance.  I therefore ...

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